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Question Number 36923 by maxmathsup by imad last updated on 07/Jun/18

for t≥0 and f(t)= (t/(√(1+t))) let S_n =Σ_(k=1) ^n f((k/n^2 )) study the convergence of S_n .

$${for}\:{t}\geqslant\mathrm{0}\:{and}\:\:{f}\left({t}\right)=\:\frac{{t}}{\sqrt{\mathrm{1}+{t}}}\:\:{let} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right)\:\:{study}\:{the}\:{convergence}\:{of}\:{S}_{{n}} \:\:. \\ $$

Commented by abdo.msup.com last updated on 10/Jun/18

we have (1+t)^α =1+αt +((α(α−1))/2)t^2 +... ⇒(1+t)^(−(1/2)) =1−(t/2) +(1/2)(−(1/2))(−(3/2))t^2 +.. =1−(t/2) +(3/8) t^2 −...⇒ 1−(t/2)≤ (1+t)^(−(1/2)) ≤ 1−(t/2) +(3/8)t^2 ∀t≥0 t−(t^2 /2)≤ f(t)≤t−(t^2 /2) +(3/8)t^3 ⇒ (k/n^2 ) −(k^2 /(2n^4 ))≤ f((k/n^2 ))≤(k/n^2 ) −(k^2 /(2n^4 )) +(3/(16)) (k^3 /n^6 ) ⇒ (1/n^2 )Σ_(k=1) ^n k−(1/(2n^4 ))Σ_(k=1) ^n k^2 ≤Σ_(k=1) ^n f((k/n^2 )) ≤ (1/n^2 )Σ_(k=1) ^n k −(1/(2n^4 ))Σ_(k=1) ^n k^2 +(3/(16n^6 ))Σ_(k=1) ^n k^3 ⇒((n(n+1))/(2n^2 )) −(1/(2n^4 ))((n(n+1)(2n+1))/6)≤ Σ_(k=1) ^n f((k/n^2 ))≤ ((n(n+1))/(2n^2 )) −((n(n+1)(2n+1))/(12n^4 )) + (3/(16n^6 )) ((n^2 (n+1)^2 )/4) ⇒ lim_(n→+∞) S_n =(1/2)

$${we}\:{have}\:\left(\mathrm{1}+{t}\right)^{\alpha} \:=\mathrm{1}+\alpha{t}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}}{t}^{\mathrm{2}} \:+... \\ $$$$\Rightarrow\left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\mathrm{1}−\frac{{t}}{\mathrm{2}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right){t}^{\mathrm{2}} +.. \\ $$$$=\mathrm{1}−\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:{t}^{\mathrm{2}} \:−...\Rightarrow \\ $$$$\mathrm{1}−\frac{{t}}{\mathrm{2}}\leqslant\:\left(\mathrm{1}+{t}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \leqslant\:\mathrm{1}−\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}{t}^{\mathrm{2}} \:\:\:\forall{t}\geqslant\mathrm{0} \\ $$$${t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\leqslant\:{f}\left({t}\right)\leqslant{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}{t}^{\mathrm{3}} \:\Rightarrow \\ $$$$\frac{{k}}{{n}^{\mathrm{2}} }\:−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\leqslant\:{f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\frac{{k}}{{n}^{\mathrm{2}} }\:−\frac{{k}^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{4}} }\:+\frac{\mathrm{3}}{\mathrm{16}}\:\frac{{k}^{\mathrm{3}} }{{n}^{\mathrm{6}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\sum_{{k}=\mathrm{1}} ^{{n}} {k}−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}^{\mathrm{2}} \leqslant\sum_{{k}=\mathrm{1}} ^{{n}} {f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$\leqslant\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{2}} \:\:+\frac{\mathrm{3}}{\mathrm{16}{n}^{\mathrm{6}} }\sum_{{k}=\mathrm{1}} ^{{n}} {k}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{4}} }\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\leqslant \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right)\leqslant\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }\:−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}{n}^{\mathrm{4}} } \\ $$$$+\:\frac{\mathrm{3}}{\mathrm{16}{n}^{\mathrm{6}} }\:\:\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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